Step 1: Group the five compounds by type.
We have one aromatic amine (aniline, $\text{C}_6\text{H}_5\text{NH}_2$), ammonia ($\text{NH}_3$), one amine with the ring separated from nitrogen by a $-\text{CH}_2-$ (benzylamine), and two aliphatic ethylamines, $\text{C}_2\text{H}_5\text{NH}_2$ and $(\text{C}_2\text{H}_5)_2\text{NH}$.
Step 2: Place aniline at the bottom.
In aniline, the nitrogen lone pair spreads into the benzene ring by resonance, so it is far less available to accept a proton. This makes aniline the weakest base of the five, weaker even than ammonia, which has no such delocalisation.
Step 3: Place benzylamine above ammonia but below the ethylamines.
In benzylamine the nitrogen is not directly joined to the ring, so there is no resonance loss, and it is a stronger base than ammonia. But the ring is still one bond away and pulls electron density slightly, so it stays weaker than simple alkyl amines.
Step 4: Order the two ethylamines using the inductive effect.
Alkyl groups push electron density onto nitrogen through the inductive effect, and a second ethyl group pushes even more, so in water $(\text{C}_2\text{H}_5)_2\text{NH}$ is a stronger base than $\text{C}_2\text{H}_5\text{NH}_2$. Putting the five together from weakest to strongest: \[ \text{C}_6\text{H}_5\text{NH}_2 < \text{NH}_3 < \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} \]
\[ \boxed{\text{C}_6\text{H}_5\text{NH}_2 < \text{NH}_3 < \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH}} \]